# Power consumption for AC motor

Discussion in 'General Electronics Chat' started by RichieG, Oct 30, 2012.

1. ### RichieG Thread Starter New Member

Dec 9, 2008
4
0
Hello all ! I am trying to evaluate the electrical power consumption of a 3 phase , 1 hp, wye connected pump motor for a piece of equipment here at work. The motor literature says that it has a power factor of 71. At the electrical control box , I measure 208 vac across each phase going out to the motor, and with a clamp on amp probe, I measure about 2 amps on each of the 3 phases. To calculate electrical power consumption, do I simply take my measurements and multiply V x I for each wire and add up the three results, or is there more to it than that? Thanks!

Oct 9, 2007
2,720
496
3. ### panic mode AAC Fanatic!

Oct 10, 2011
1,766
533
certainly using wattmeter is the easiest way. if you want to calculate power then:

Pdc=V*I
Pac1=V*I*cos(θ)
Pac3=sqrt(3)*VLL*I*cos(θ)
where cos(θ) is phase difference between V and I. you may use value stamped on the motor. note it is 0.71, not 71.

If you measured voltages phase to neutral (Y circuit) then
Pac3=3*VLN*I*cos(θ)

because VLL = VLN*sqrt(3) in a 3-phase circuit
VLL= Voltage Line to Line (phase to phase)
VLN = Voltage Line to Neutral

4. ### RichieG Thread Starter New Member

Dec 9, 2008
4
0
Thanks for all your help. Since this is just for a ballpark estimate and don't have a watt meter at my disposal, I'll just go with calculations based on my readings.

5. ### Dodgydave AAC Fanatic!

Jun 22, 2012
8,273
1,444
3phase power calculations

Total power drawn from supply W = V x I x 1.732

actual load power given out W = V x I x 1.732 x PF(power factor)

so in your case 208V x 2Amps x 1.732= 720.512W actual power drawn from supply.
Motor power given out is 720.512 x 0.71 = 511.56W

so the higher to a power factor of 1 is better as your wasting money!!!